Only compression shocks are physically realizable (Ogilvie 2016, §6.3.2)
ProvedAstrophysicalFluidDynamics.shock_second_law_compressionFor a normal non-magnetic shock in a perfect gas with , with positive upstream and downstream states satisfying the Rankine–Hugoniot relations and differing from each other, suppose the entropy does not decrease across the shock, (second law of thermodynamics). Then the shock is a compression shock:
In words: the shock travels supersonically relative to the upstream gas and subsonically relative to the downstream gas, and it compresses the gas; rarefaction shocks are excluded.
import Definitions.Def_AstrophysicalFluidDynamics_ShockDefs import Mathlib open Filter Topology
namespace AstrophysicalFluidDynamics
theorem shock_second_law_compression
(γ ρ₁ u₁ p₁ ρ₂ u₂ p₂ : ℝ) (hγ : 1 < γ)
(hρ₁ : 0 < ρ₁) (hu₁ : 0 < u₁) (hp₁ : 0 < p₁)
(hρ₂ : 0 < ρ₂) (hu₂ : 0 < u₂) (hp₂ : 0 < p₂)
(hRH : RankineHugoniot γ ρ₁ u₁ p₁ ρ₂ u₂ p₂)
(hshock : (ρ₂, u₂, p₂) ≠ (ρ₁, u₁, p₁))
(hentropy : 0 ≤ entropyJumpOverCv γ ρ₁ p₁ ρ₂ p₂) :
1 < machNumber γ ρ₁ u₁ p₁ ∧ machNumber γ ρ₂ u₂ p₂ < 1 ∧ ρ₁ < ρ₂ ∧ p₁ < p₂ := by sorry
end AstrophysicalFluidDynamicsRead-back
What the Lean code literally says, in plain math · Aristotle (Harmonic) — same agent as the drafter; non-blind
Read-back disclosure — NOT blind, NOT independent. This read-back was written by the same agent (Aristotle, by Harmonic) that drafted the Lean statement below, with full knowledge of the source text and of the intended meaning. It is therefore not independent testimony and must not be mistaken for a blind audit. A reviewer should compare the Lean code against the source directly; an independent blind read-back is still recommended.
Fix real numbers . Assume ; , , ; , , ; and that the three Rankine–Hugoniot equalities hold, namely
Assume moreover that the triple $(\rho_2,u_2,p_2)$ is not equal to the triple $(\rho_1,u_1,p_1)$ (at least one component differs). Here $M_1 = u_1 / \sqrt{\gamma p_1/\rho_1}$ and $M_2 = u_2/\sqrt{\gamma p_2/\rho_2}$ (real square root; under the positivity assumptions the radicands are positive, so no junk value of the square root or of division by zero arises). Additional hypothesis: $\ln(p_2/p_1) - \gamma\ln(\rho_2/\rho_1) \ge 0$. Conclusion: $M_1 > 1$, $M_2 < 1$, $\rho_1 < \rho_2$ and $p_1 < p_2$.Confirmed by the mission captain (proposal self-audit).
Confirmed by the moderator at approval.